package leetcode.editor.cn;//请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果
//一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。 
//
// [["a","b","c","e"], 
//["s","f","c","s"], 
//["a","d","e","e"]] 
//
// 但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。 
//
// 
//
// 示例 1： 
//
// 
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "AB
//CCED"
//输出：true
// 
//
// 示例 2： 
//
// 
//输入：board = [["a","b"],["c","d"]], word = "abcd"
//输出：false
// 
//
// 
//
// 提示： 
//
// 
// 1 <= board.length <= 200 
// 1 <= board[i].length <= 200 
// 
//
// 注意：本题与主站 79 题相同：https://leetcode-cn.com/problems/word-search/ 
// Related Topics 深度优先搜索 
// 👍 246 👎 0


//leetcode submit region begin(Prohibit modification and deletion)
class SolutionJZ12 {
    public boolean exist(char[][] board, String word) {
        if (board == null || board[0] == null || board.length == 0 || board[0].length == 0 || word == null || word.equals("")) {
            return false;
        }
        boolean[][] isVisited = new boolean[board.length][board[0].length];
        char[] chs = word.toCharArray();

        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == chs[0]) {
                    if (bfs(board, i, j, isVisited, chs, 0)) return true;
                }
            }
        }
        return false;

    }

    private boolean bfs(char[][] board, int i, int j, boolean[][] isVisited, char[] chs, int index) {

        if (index == chs.length) {
            return true;
        }
        if (i < 0 || j < 0 || i == board.length || j == board[0].length || isVisited[i][j] || board[i][j] != chs[index]) {
            return false;
        }
        isVisited[i][j] = true;
        boolean res = bfs(board, i + 1, j, isVisited, chs, index + 1)
                || bfs(board, i - 1, j, isVisited, chs, index + 1)
                || bfs(board, i, j + 1, isVisited, chs, index + 1)
                || bfs(board, i, j - 1, isVisited, chs, index + 1);
        isVisited[i][j] = false;
        return res;
    }
}
//leetcode submit region end(Prohibit modification and deletion)
